What is the distribution of an unconditioned random variable knowing the conditional distribution? I have two random variables $X$ and $Y$. I know that $Y$ can be approximated by a $N(\mu_1,\sigma_1^2)$ distribution (in particular $Y$ is not negative) and I also know that $X|Y \sim N(a+bY,c+dY)$ for some constants $a,b,c,d$. What can I infer about the distribution of $X$? Is there some simple formula I can derive? My intuition tells me that it might also a normal distribution with $\mu_2 = a+b\mu_1$ but I can't tell what the variance will be. I'm not sure that this will be the case though. If anyone knows of a closed form formula that would be of tremendous help.

The distribution of $X$ will be given by \begin{align} f(x) &=\int_{-\infty}^\infty f(x,y)dy\\\ &=\int_{-\infty}^\infty f(x|y) f(y) dy\\\ &=\int_{-\infty}^\infty \frac{e^{-(x-a-by)^2/(c+dy)}}{\sqrt{2\pi(c+dy)}} \frac{e^{-(y-\mu_1)^2/\sigma_1^2}}{\sqrt{2\pi}\sigma_1} dy\\\ \end{align} The problem with the integral is that you have a mixed exponent ($2bxy$) in the first exponential that won't allow you to factor out a function of $x$ as a factor and have an integrand that is independent of $x$. To find an analytical expression for the integral looks infeasible to me. Maybe ask mathematica or something similar for that.

BTW: Are you sure about your conditional probability? With finite probability you have a negative variance for X!